Question

Consider the titration of a 28.0 −mL sample of 0.170 M CH3NH2 with 0.145 M HBr. Determine each of the following.

a) the initial ph

b)the volume of added acid required to reach the equivalence point

c)the pH at 4.0 mL of added acid

d)the pH at one-half of the equivalence point

e)the pH at the equivalence point

f)the pH after adding 6.0 mL of acid beyond the equivalence point

Answer 1

Given that concentration of HBr = 0.145 M concentration of CH3NH2 = 0.170 M Volume of CH3NH2 = 28.0 mL Kb of CH3NH, = 4.4 x 10-4 a) Before Addition of any acid: M CH3NH2 + H2O CH3NH2+ + Initial 0.170 Change Equillibrium 0.170 - X K =_X.X =4.4x10-4 0.170- x x +4.4x10+ – 7.48x10-4 = 0' by solving we get x = 0.008431 :: [OH"]=0.008431M pOH =-log(0.008431)=2.07 pH = 14 – 2.07 = 11.93

b) At equivalence point: Number of moles of CH3NH, = M*V = 0.170 M* 28.0 mL = 4.76 mmol At equivalence point, the number of moles CH3NH, of is equal to the number of moles of HBr. Number of moles of HBr = 4.76 mmol Volume of HBr required to reach equivalence point = moles/molarity = 4.76 mmol/0.145 M = 32.83 mL Volume of acid required to reach equivalence point = 32.83 m

c) Addition of 4.0 mL of HBr: Number of moles of CH3NH, = M*V = 0.170 M* 28.0 mL = 4.76 mmol Number of moles of HBr = M*V = 0.145 M* 4.0 mL = 0.58 mmol Net moles of CH3NH2 = 4.76 -0.58 = 4.18 mmol This is the mid point. mmol CH3NH2 + HBr = CH3NH2+Br Initial 0 4.76 -0.58 Change -0.58 +0.58 Equillibrium 4.18 0 0.58 According to Henderson's equation ſacid] 0.58 pOH = pK, +log T= -log(4.4x10) + log =3.36-0.85775 [base] 4.18 pOH = 2.50 pH = 14- pOH = 14 -2.50 = 11.50 (d) At half equivalence point, pOH = pkb = 3.36 pH = 14 - 3.36 = 10.64

(e) At this point, there is no CH3NH, is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles 4.76 mmol 4.701 THNA; B1 = 0.078254M Total volume (28.0 + 32.83) mL K, of CHẶNH,"Br"] = = 1x10 = 2.27x10-11 Dissociation of CH3NH3 + is given by mmol CH2NH3 - CHỌNH + Initial 0.078254 Change -X Equillibrium 0.078254 - X K, = 2.27x10-11 0.078254-X x is small, so 0.078254- x=0.078254 20.078254 = 2.27x10-11 x2 = 1.78x10-12 >x=1.33x10^M :: [H*] =1.33x10^M pH = -log (1.33x10-6) = 5.87

f) Addition of 6 ml beyond equi. Point (32.83 +6.00 = 38.83 ) 38.83 mL of HBr: Number of moles of CH NH, = M*V = 0.170 M* 28.0 mL = 4.76 mmol Number of moles of HBr = M*V = 0.145 M* 38.83 mL = 5.63 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HBr in the solution. Net moles of HBr = 5.63 – 4.76 = 0.87 mmol Total volume of the solution = 28 + 38.83 = 66.83 mL Conc. of HBr = [H+] = moles/ volume = 0.87 mmol / 66.83 mL = 0.013019 M pH =- log [H+] = -log[0.013019] = 1.89